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<P align=left><FONT color=#ff9933><B><FONT color=#3300cc
size=5>经典c程序100例==71--80</FONT></B></FONT></P>
<P><FONT
color=#990000>【程序71】</FONT><BR>题目：编写input()和output()函数输入，输出5个学生的数据记录。<BR>1.程序分析：<BR>2.程序源代码：<BR>#define
N 5<BR>struct student<BR>{ char num[6];<BR> char name[8];<BR> int
score[4];<BR>} stu[N];<BR>input(stu)<BR>struct student stu[];<BR>{ int
i,j;<BR> for(i=0;i<N;i++)<BR> { printf("\n please input %d of
%d\n",i+1,N);<BR> printf("num:
");<BR> scanf("%s",stu[i].num);<BR> printf("name:
");<BR> scanf("%s",stu[i].name);<BR> for(j=0;j<3;j++)<BR> {
printf("score
%d.",j+1);<BR> scanf("%d",&stu[i].score[j]);<BR> }<BR> printf("\n");<BR> }<BR>}<BR>print(stu)<BR>struct
student stu[];<BR>{ int i,j;<BR>printf("\nNo. Name Sco1 Sco2
Sco3\n");<BR>for(i=0;i<N;i++)<BR>{
printf("%-6s%-10s",stu[i].num,stu[i].name);<BR> for(j=0;j<3;j++)<BR> printf("%-8d",stu[i].score[j]);<BR> printf("\n");<BR>}<BR>}<BR>main()<BR>{<BR> input();<BR> print();<BR>}<BR>==============================================================<BR><FONT
color=#990000>【程序72】</FONT><BR>题目：创建一个链表。<BR>1.程序分析： <BR>2.程序源代码：<BR>/*creat
a list*/<BR>#include "stdlib.h"<BR>#include "stdio.h"<BR>struct list<BR>{
int data;<BR>struct list *next;<BR>};<BR>typedef struct list
node;<BR>typedef node *link;<BR>void main()<BR>{ link ptr,head;<BR>int
num,i;<BR>ptr=(link)malloc(sizeof(node));<BR>ptr=head;<BR>printf("please
input 5
numbers==>\n");<BR>for(i=0;i<=4;i++)<BR>{<BR> scanf("%d",&num);<BR> ptr->data=num;<BR> ptr->next=(link)malloc(sizeof(node));<BR> if(i==4)
ptr->next=NULL;<BR> else
ptr=ptr->next;<BR>}<BR>ptr=head;<BR>while(ptr!=NULL)<BR>{ printf("The
value is
==>%d\n",ptr->data);<BR> ptr=ptr->next;<BR>}<BR>}<BR>==============================================================<BR><FONT
color=#990000>【程序73】</FONT><BR>题目：反向输出一个链表。 <BR>1.程序分析：<BR>2.程序源代码：<BR>/*reverse
output a list*/<BR>#include "stdlib.h"<BR>#include "stdio.h"<BR>struct
list<BR>{ int data;<BR> struct list *next;<BR>};<BR>typedef struct list
node;<BR>typedef node *link;<BR>void main()<BR>{ link
ptr,head,tail; <BR> int
num,i;<BR> tail=(link)malloc(sizeof(node));<BR> tail->next=NULL;<BR> ptr=tail;<BR> printf("\nplease
input 5
data==>\n");<BR> for(i=0;i<=4;i++)<BR> {<BR> scanf("%d",&num);<BR> ptr->data=num;<BR> head=(link)malloc(sizeof(node));<BR> head->next=ptr;<BR> ptr=head;<BR> }<BR>ptr=ptr->next;<BR>while(ptr!=NULL)<BR>{
printf("The value is
==>%d\n",ptr->data);<BR> ptr=ptr->next;<BR>}}<BR>==============================================================<BR><FONT
color=#990000>【程序74】</FONT><BR>题目：连接两个链表。<BR>1.程序分析：<BR>2.程序源代码：<BR>#include
"stdlib.h"<BR>#include "stdio.h"<BR>struct list<BR>{ int data;<BR>struct
list *next;<BR>};<BR>typedef struct list node;<BR>typedef node
*link;<BR>link delete_node(link pointer,link tmp)<BR>{if (tmp==NULL)
/*delete first node*/<BR> return pointer->next;<BR>else<BR>{
if(tmp->next->next==NULL)/*delete last
node*/<BR> tmp->next=NULL;<BR> else /*delete the other
node*/<BR> tmp->next=tmp->next->next;<BR> return
pointer;<BR>}<BR>}<BR>void selection_sort(link pointer,int num)<BR>{ link
tmp,btmp;<BR> int
i,min;<BR> for(i=0;i<num;i++)<BR> {<BR> tmp=pointer;<BR> min=tmp->data;<BR> btmp=NULL;<BR> while(tmp->next)<BR> {
if(min>tmp->next->data)<BR> {min=tmp->next->data;<BR> btmp=tmp;<BR> }<BR> tmp=tmp->next;<BR> }<BR>printf("\40:
%d\n",min);<BR>pointer=delete_node(pointer,btmp);<BR>}<BR>}<BR>link
create_list(int array[],int num)<BR>{ link tmp1,tmp2,pointer;<BR>int
i;<BR>pointer=(link)malloc(sizeof(node));<BR>pointer->data=array[0];<BR>tmp1=pointer;<BR>for(i=1;i<num;i++)<BR>{
tmp2=(link)malloc(sizeof(node));<BR> tmp2->next=NULL;<BR> tmp2->data=array[i];<BR> tmp1->next=tmp2;<BR> tmp1=tmp1->next;<BR>}<BR>return
pointer;<BR>}<BR>link concatenate(link pointer1,link pointer2)<BR>{ link
tmp;<BR>tmp=pointer1;<BR>while(tmp->next)<BR> tmp=tmp->next;<BR>tmp->next=pointer2;<BR>return
pointer1;<BR>}<BR>void main(void)<BR>{ int arr1[]={3,12,8,9,11};<BR> link
ptr;<BR> ptr=create_list(arr1,5);<BR> selection_sort(ptr,5);<BR>}<BR>==============================================================<BR><FONT
color=#990000>【程序75】</FONT><BR>题目：放松一下，算一道简单的题目。<BR>1.程序分析：<BR>2.程序源代码：<BR>main()<BR>{<BR>int
i,n;<BR>for(i=1;i<5;i++)<BR>{
n=0;<BR> if(i!=1)<BR> n=n+1;<BR> if(i==3)<BR> n=n+1;<BR> if(i==4)<BR> n=n+1;<BR> if(i!=4)<BR> n=n+1;<BR> if(n==3)<BR> printf("zhu
hao shi de
shi:%c",64+i);<BR> }<BR>}<BR>==============================================================<BR><FONT
color=#990000>【程序76】</FONT><BR>题目：编写一个函数，输入n为偶数时，调用函数求1/2+1/4+...+1/n,当输入n为奇数时，调用函数<BR> 1/1+1/3+...+1/n(利用指针函数)<BR>1.程序分析：<BR>2.程序源代码：<BR>main()<BR>#include
"stdio.h"<BR>main()<BR>{<BR>float peven(),podd(),dcall();<BR>float
sum;<BR>int n;<BR>while
(1)<BR>{<BR> scanf("%d",&n);<BR> if(n>1)<BR> break;<BR>}<BR>if(n%2==0)<BR>{<BR> printf("Even=");<BR> sum=dcall(peven,n);<BR>}<BR>else<BR>{<BR> printf("Odd=");<BR> sum=dcall(podd,n);<BR>}<BR>printf("%f",sum);<BR>}<BR>float
peven(int n)<BR>{<BR>float s;<BR>int
i;<BR>s=1;<BR>for(i=2;i<=n;i+=2)<BR> s+=1/(float)i;<BR>return(s);<BR>}<BR>float
podd(n)<BR>int n;<BR>{<BR>float s;<BR>int
i;<BR>s=0;<BR>for(i=1;i<=n;i+=2)<BR> s+=1/(float)i;<BR>return(s);<BR>}<BR>float
dcall(fp,n)<BR>float (*fp)();<BR>int n;<BR>{<BR>float
s;<BR>s=(*fp)(n);<BR>return(s);<BR>}<BR>==============================================================<BR><FONT
color=#990000>【程序77】</FONT><BR>题目：填空练习（指向指针的指针）<BR>1.程序分析： <BR>2.程序源代码：<BR>main()<BR>{
char *s[]={"man","woman","girl","boy","sister"};<BR>char **q;<BR>int
k;<BR>for(k=0;k<5;k++)<BR>{ ;/*这里填写什么语句*/<BR> printf("%s\n",*q);<BR>}<BR>}<BR>==============================================================<BR><FONT
color=#990000>【程序78】</FONT><BR>题目：找到年龄最大的人，并输出。请找出程序中有什么问题。<BR>1.程序分析：<BR>2.程序源代码：<BR>#define
N 4<BR>#include "stdio.h"<BR>static struct man<BR>{ char name[20];<BR>int
age;<BR>}
person[N]={"li",18,"wang",19,"zhang",20,"sun",22};<BR>main()<BR>{struct
man *q,*p;<BR>int i,m=0;<BR>p=person;<BR>for
(i=0;i<N;i++)<BR>{if(m<p->age)<BR> q=p++;<BR> m=q->age;}<BR>printf("%s,%d",(*q).name,(*q).age);<BR>}<BR>==============================================================<BR><FONT
color=#990000>【程序79】</FONT><BR>题目：字符串排序。<BR>1.程序分析：<BR>2.程序源代码：<BR>main()<BR>{<BR>char
*str1[20],*str2[20],*str3[20];<BR>char swap();<BR>printf("please input
three
strings\n");<BR>scanf("%s",str1);<BR>scanf("%s",str2);<BR>scanf("%s",str3);<BR>if(strcmp(str1,str2)>0)
swap(str1,str2);<BR>if(strcmp(str1,str3)>0)
swap(str1,str3);<BR>if(strcmp(str2,str3)>0)
swap(str2,str3);<BR>printf("after being
sorted\n");<BR>printf("%s\n%s\n%s\n",str1,str2,str3);<BR>}<BR>char
swap(p1,p2)<BR>char *p1,*p2;<BR>{<BR>char
*p[20];<BR>strcpy(p,p1);strcpy(p1,p2);strcpy(p2,p);<BR>}<BR>==============================================================<BR><FONT
color=#990000>【程序80】</FONT><BR>题目：海滩上有一堆桃子，五只猴子来分。第一只猴子把这堆桃子凭据分为五份，多了一个，这只<BR> 猴子把多的一个扔入海中，拿走了一份。第二只猴子把剩下的桃子又平均分成五份，又多了<BR> 一个，它同样把多的一个扔入海中，拿走了一份，第三、第四、第五只猴子都是这样做的，<BR> 问海滩上原来最少有多少个桃子？<BR>1.程序分析：<BR>2.程序源代码：<BR>main()<BR>{int
i,m,j,k,count;<BR>for(i=4;i<10000;i+=4)<BR>{
count=0;<BR>m=i;<BR>for(k=0;k<5;k++)<BR>{<BR> j=i/4*5+1;<BR> i=j;<BR> if(j%4==0)<BR> count++;<BR> else<BR> break;<BR>}<BR> i=m;<BR> if(count==4)<BR> {printf("%d\n",count);<BR> break;}<BR>}<BR>}</P>
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